3 ms·
I know it's from the article, but the expression "Accelerate Electrons just below the speed of light" tells you pretty much nothing. An electron travelling at .
by archgoon 9y ago
I know it's from the article, but the expression "Accelerate Electrons just below the speed of light" tells you pretty much nothing. An electron travelling at .9 the speed of light has 1.1 MeV of energy.
Nature seems to be down, so I can't access the abstract, but the google tagline has the following.
'Photons emitted above 100 MeV as a function of target width, for 1 × 109 incident electrons of energy 500 MeV (blue), 1 GeV (black) and 2 GeV (red).'
A table for comparison
.9 1.1 MeV
.99 3.6 MeV
.999 11 MeV
.9999 36 MeV
.99999 114 MeV
.999999 361 MeV
.9999999 1.14 GeV
So it's somewhere in the .999999c-.999999999c. Assuming that snippet was accurate. The LHC gets particles up to the TeV range, but, as the name suggests, they are much larger particles. An actual particle physicist can probably tell you more about whether it is significantly harder to get the much lighter Electron (and associated radiation losses) up to those energies.
SLAC (https://en.wikipedia.org/wiki/SLAC_National_Accelerator_Laboratory https://en.wikipedia.org/wiki/SLAC_National_Accelerator_Labo...) apparently can get electrons up to 50 GeV, so if those energies from the snippet are the target ones, that part should be achievable.
- PoachedSausage 9y agoI'm not a particle physicist but a particle accelerator engineer. GeV energies for electrons are achievable with a reasonably sized synchrotron like Diamond Light Source[0]. The magnets required to keep electrons in line are quite a bit smaller than those required for protons due to the mass difference. Small refrigerator sized verses small car sized. [0] https://en.wikipedia.org/wiki/Diamond_Light_Source https://en.wikipedia.org/wiki/Diamond_Light_Source